結果

問題 No.1078 I love Matrix Construction
ユーザー stoq
提出日時 2020-12-04 19:18:31
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 67 ms / 2,000 ms
コード長 6,831 bytes
コンパイル時間 3,235 ms
コンパイル使用メモリ 218,964 KB
最終ジャッジ日時 2025-01-16 15:59:47
ジャッジサーバーID
(参考情報)
judge5 / judge3
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
other AC * 22
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#define MOD_TYPE 1
#pragma region Macros
#include <bits/stdc++.h>
using namespace std;
#if 0
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif
#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif
using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;
constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr double PI = acos(-1.0);
constexpr double EPS = 1e-7;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};
#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";
struct io_init
{
io_init()
{
cin.tie(0);
ios::sync_with_stdio(false);
cout << setprecision(30) << setiosflags(ios::fixed);
};
} io_init;
template <typename T>
inline bool chmin(T &a, T b)
{
if (a > b)
{
a = b;
return true;
}
return false;
}
template <typename T>
inline bool chmax(T &a, T b)
{
if (a < b)
{
a = b;
return true;
}
return false;
}
inline ll CEIL(ll a, ll b)
{
return (a + b - 1) / b;
}
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val)
{
fill((T *)array, (T *)(array + N), val);
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept
{
is >> p.first >> p.second;
return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> &p) noexcept
{
os << p.first << " " << p.second;
return os;
}
#pragma endregion
namespace atcoder
{
namespace internal
{
template <class E>
struct csr
{
std::vector<int> start;
std::vector<E> elist;
csr(int n, const std::vector<std::pair<int, E>> &edges)
: start(n + 1), elist(edges.size())
{
for (auto e : edges)
{
start[e.first + 1]++;
}
for (int i = 1; i <= n; i++)
{
start[i] += start[i - 1];
}
auto counter = start;
for (auto e : edges)
{
elist[counter[e.first]++] = e.second;
}
}
};
// Reference:
// R. Tarjan,
// Depth-First Search and Linear Graph Algorithms
struct scc_graph
{
public:
scc_graph(int n) : _n(n) {}
int num_vertices() { return _n; }
void add_edge(int from, int to) { edges.push_back({from, {to}}); }
// @return pair of (# of scc, scc id)
std::pair<int, std::vector<int>> scc_ids()
{
auto g = csr<edge>(_n, edges);
int now_ord = 0, group_num = 0;
std::vector<int> visited, low(_n), ord(_n, -1), ids(_n);
visited.reserve(_n);
auto dfs = [&](auto self, int v) -> void {
low[v] = ord[v] = now_ord++;
visited.push_back(v);
for (int i = g.start[v]; i < g.start[v + 1]; i++)
{
auto to = g.elist[i].to;
if (ord[to] == -1)
{
self(self, to);
low[v] = std::min(low[v], low[to]);
}
else
{
low[v] = std::min(low[v], ord[to]);
}
}
if (low[v] == ord[v])
{
while (true)
{
int u = visited.back();
visited.pop_back();
ord[u] = _n;
ids[u] = group_num;
if (u == v)
break;
}
group_num++;
}
};
for (int i = 0; i < _n; i++)
{
if (ord[i] == -1)
dfs(dfs, i);
}
for (auto &x : ids)
{
x = group_num - 1 - x;
}
return {group_num, ids};
}
std::vector<std::vector<int>> scc()
{
auto ids = scc_ids();
int group_num = ids.first;
std::vector<int> counts(group_num);
for (auto x : ids.second)
counts[x]++;
std::vector<std::vector<int>> groups(ids.first);
for (int i = 0; i < group_num; i++)
{
groups[i].reserve(counts[i]);
}
for (int i = 0; i < _n; i++)
{
groups[ids.second[i]].push_back(i);
}
return groups;
}
private:
int _n;
struct edge
{
int to;
};
std::vector<std::pair<int, edge>> edges;
};
} // namespace internal
} // namespace atcoder
namespace atcoder
{
// Reference:
// B. Aspvall, M. Plass, and R. Tarjan,
// A Linear-Time Algorithm for Testing the Truth of Certain Quantified Boolean
// Formulas
struct two_sat
{
public:
two_sat() : _n(0), scc(0) {}
two_sat(int n) : _n(n), _answer(n), scc(2 * n) {}
void add_clause(int i, bool f, int j, bool g)
{
assert(0 <= i && i < _n);
assert(0 <= j && j < _n);
scc.add_edge(2 * i + (f ? 0 : 1), 2 * j + (g ? 1 : 0));
scc.add_edge(2 * j + (g ? 0 : 1), 2 * i + (f ? 1 : 0));
}
bool satisfiable()
{
auto id = scc.scc_ids().second;
for (int i = 0; i < _n; i++)
{
if (id[2 * i] == id[2 * i + 1])
return false;
_answer[i] = id[2 * i] < id[2 * i + 1];
}
return true;
}
std::vector<bool> answer() { return _answer; }
private:
int _n;
std::vector<bool> _answer;
internal::scc_graph scc;
};
} // namespace atcoder
void solve()
{
int n;
cin >> n;
vector<int> s(n), t(n), u(n);
rep(i, n) cin >> s[i], s[i]--;
rep(i, n) cin >> t[i], t[i]--;
rep(i, n) cin >> u[i];
atcoder::two_sat ts(n * n);
rep(i, n)
{
bool a = !(u[i] & 1), b = !(u[i] & 2);
rep(j, n)
{
int p = s[i] * n + j, q = j * n + t[i];
ts.add_clause(p, a, q, b);
}
}
bool ex = ts.satisfiable();
if (!ex)
{
cout << -1 << "\n";
return;
}
auto ans = ts.answer();
rep(i, n * n)
{
cout << ans[i] << (i % n == n - 1 ? "\n" : " ");
}
}
int main()
{
solve();
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0